Termination w.r.t. Q proof of TRS/secret06matchbox-gen-25.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

Q restricted rewrite system:
The TRS R consists of the following rules:

b₂(a, f₁(b₂(b₂(z, y), a))) -> z
c₃(c₃(z, x, a), a, y) -> f₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
f₁(f₁(c₃(a, y, z))) -> b₂(y, b₂(z, z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

b₂(a, f₁(b₂(b₂(z, y), a))) -> z
c₃(c₃(z, x, a), a, y) -> f₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
f₁(f₁(c₃(a, y, z))) -> b₂(y, b₂(z, z))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F₁(f₁(c₃(a, y, z))) -> B₂(y, b₂(z, z))
C₃(c₃(z, x, a), a, y) -> F₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
C₃(c₃(z, x, a), a, y) -> F₁(c₃(z, y, x))
F₁(f₁(c₃(a, y, z))) -> B₂(z, z)
C₃(c₃(z, x, a), a, y) -> F₁(c₃(y, a, f₁(c₃(z, y, x))))
C₃(c₃(z, x, a), a, y) -> C₃(z, y, x)
C₃(c₃(z, x, a), a, y) -> C₃(y, a, f₁(c₃(z, y, x)))

The TRS R consists of the following rules:

b₂(a, f₁(b₂(b₂(z, y), a))) -> z
c₃(c₃(z, x, a), a, y) -> f₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
f₁(f₁(c₃(a, y, z))) -> b₂(y, b₂(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

F₁(f₁(c₃(a, y, z))) -> B₂(y, b₂(z, z))
C₃(c₃(z, x, a), a, y) -> F₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
C₃(c₃(z, x, a), a, y) -> F₁(c₃(z, y, x))
F₁(f₁(c₃(a, y, z))) -> B₂(z, z)
C₃(c₃(z, x, a), a, y) -> F₁(c₃(y, a, f₁(c₃(z, y, x))))
C₃(c₃(z, x, a), a, y) -> C₃(z, y, x)
C₃(c₃(z, x, a), a, y) -> C₃(y, a, f₁(c₃(z, y, x)))

The TRS R consists of the following rules:

b₂(a, f₁(b₂(b₂(z, y), a))) -> z
c₃(c₃(z, x, a), a, y) -> f₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
f₁(f₁(c₃(a, y, z))) -> b₂(y, b₂(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

Q DP problem:
The TRS P consists of the following rules:

C₃(c₃(z, x, a), a, y) -> C₃(y, a, f₁(c₃(z, y, x)))
C₃(c₃(z, x, a), a, y) -> C₃(z, y, x)

The TRS R consists of the following rules:

b₂(a, f₁(b₂(b₂(z, y), a))) -> z
c₃(c₃(z, x, a), a, y) -> f₁(f₁(c₃(y, a, f₁(c₃(z, y, x)))))
f₁(f₁(c₃(a, y, z))) -> b₂(y, b₂(z, z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.